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ENCE 4610Foundation Analysis and
Design
Shallow Foundations: Part I
Topics for Shallow Foundations, Part I
● Types of Shallow Foundations
● Spread Footing Design Concept and Procedure
● Bearing Capacity Failure Mechanisms Bearing Capacity
Equation Formulation Bearing Capacity
Correction Factors
● Other Items Local or Punching Shear Factors of Safety Practical Aspects of
Bearing Capacity Formulations
Presumptive Bearing Capacities
Types of Shallow Foundations● Shallow foundations are usually
placed within a depth D beneath the ground surface less than the minimum width B of the foundation
● Shallow foundations consist of:– Spread and continuous footings– Square, Rectangular or Circular
Footings– Continuous footings– Ring Foundations– Strap Footings
● Wall footings● Mats or Rafts
Footings A finite spread footing
is a shallow foundation that transmits loads and has an aspect ratio of 1 < L/B < 10
A continuous spread footing is an “infinite” footing where L/B > 10 and the effects of L are ignored
Abutment/Wingwall Footing● A situation where the
shallow foundation is a part of and acts with a retaining wall for both vertical load-bearing and horizontal loads of retained soil
Combined Footing● Combined footings are similar to
isolated spread footings except that they support two or more columns and are rectangular or trapezoidal in shape (Figure 8-7). They are used primarily when the column spacing is non-uniform (Bowles, 1996) or when isolated spread footings become so closely spaced that a combination footing is simpler to form and construct. In the case of bridge abutments, an example of a combined footing is the so-called “spill-through” type abutment (Figure 8-8). This configuration was used during some of the initial construction of the Interstate Highway System on new alignments where spread footings could be founded on competent native soils. Spill-through abutments are also used at stream crossings to make sure that foundations are below the scour depth of the stream.
Mat FoundationsMat Foundations A mat is continuous in two
directions capable of supporting multiple columns, wall or floor loads. It has dimensions from 20 to 80 ft or more for houses and hundreds of feet for large structures such as multi-story hospitals and some warehouses
Ribbed mats, consisting of stiffening beams placed below a flat slab are useful in unstable soils such as expansive, collapsible or soft materials where differential movements can be significant (exceeding 0.5 inch).
Bearing Pressure DistributionBearing Pressure DistributionConcentric LoadsConcentric Loads
Flexible foundation on clay
Flexible Foundation on Sand
Rigid foundation on clay
Rigid Foundation on Sand
Simplified Distribution
Shear Failure vs. Settlement in Allowable Bearing Capacity
Plasticity: Lower and Upper Bound Solutions
● The Problem– Bearing Capacity failure is a plastic failure of the soil
along some failure surface– The problem of plastic failure of twofold:
● Finding the failure surface along which the plasticity an ultimately failure takes place
● Determining the failure state to which we should design, i.e. lower or upper bound
● The first is done by determining which failure surface provides the “path of least resistance” of failure
● The second is in part driven by uncertainty requirements in failure
● Review of Upper and Lower Bound Concept– Lower Bound: The true failure load is larger than the
load corresponding to an equilibrium system. The system has failed in at least one place.
– Upper Bound: The true failure load is smaller than the load corresponding to a mechanism, if that load is determined using the virtual work principle. The system has failed “in general.”
– The idea is that the true solution is somewhere between the two
● We saw this when we went through unsupported cuts in purely cohesive soils– In that case, we had to consider both the shape of
the failure surface and its location– For slopes, a circular failure surface was
considered as the most likely failure surface– The actual surface could be located for simple
slopes using theoretical considerations, but for more complex slopes (layered soils, water table, frictional soils) a trial and error solution was adopted
● In principle, only applicable to purely cohesive soils without friction, due to volume expansion considerations– Upper and lower bound theory can be extended to
soils with a frictional component (or only a frictional component,) but the implementation is much more complicated
● We will begin by considering strip (infinite or continuous) foundations only) in cohesive soils
Lower Bound Solution
By direct application and Mohr's Circle
By theory of elasticity
The more realistic lower bound
Upper Bound Limit Equilibrium MethodUpper Bound Limit Equilibrium Method
(Circular Failure Surface, Cohesive Soil)(Circular Failure Surface, Cohesive Soil)0
22 0=
BBbσBcBbπcBbBBbq=M ultp
Assume: No soil strength due to internal friction (cohesive soil,) shear strength above foundation base neglectedWe add the effect of the weight of the soil (effective stress) acting on the top of the right
side of the circle against rotation.
0cult
c
0ult
σ+cN=qπcBb=N
σ+cπcBb=q6.282
2
More Realistic Upper Bound Case● This is done by moving
the centre of the failure surface upward
● A circular failure surface is what we assumed for slope failure
● It is valid for very soft clays, and methods have been developed for use with these types of soils
● Soft clays are more subject to settlement
● We will not use these in this course
So the solution is bounded by
Development of Prandtl Bearing Capacity Theory
● Application of limit equilibrium methods first done by Prandtl on the punching of thick masses of metal (materials with no internal frictional effects)
● Prandtl's methods first adapted by Terzaghi to bearing capacity failure of shallow foundations (specifically, he added the effects of frictional materials)
● Vesić and others (Meyerhof, Brinch Hansen, etc.) improved on Terzaghi's original theory and added other factors for a more complete analysis
Note the three zones, the foundation fails along the lower boundary of these zones
Assumptions for Bearing Assumptions for Bearing Capacity MethodsCapacity Methods
Geometric assumption– Depth of foundation is less than
or equal to its width– Foundation is a strip footing
(infinite length)*
Geotechnical Assumptions– Soil beneath foundation is
homogeneous semi-infinite mass*– Mohr-Coulomb model for soil– General shear failure mode is the
governing mode*– No soil consolidation occurs– Soil above bottom of foundation
has no shear strength; is only a surcharge load against the overturning load*
Foundation-Soil Interface Assumptions Foundation is very rigid
relative to the soil No sliding occurs
between foundation and soil (rough foundation)
Loading Assumptions Applied load is
compressive and applied vertically to the centroid of the foundation*
No applied moments present*
* We will discuss “workarounds” to these assumptions
Loads and Failure Zones Loads and Failure Zones on Strip Foundationson Strip Foundations
Q = load/unit length on foundationq = load/unit area due to effective stress at base of foundationI, II, III = regions of failure in Prantdl theory for bearing capacity failure
Basic Equation of Bearing Capacity
Basic Equation of Bearing Capacity
● Values of Nc, Nq mostly the same. Values of Nγ depend upon theory
● DIN/Brinch-Hansen:
● CFEM:
● Vesić:
• Vesić-AASHTO Factors:
Bearing Capacity Example
Types of Bearing Capacity Traditionally, bearing capacity has
been classified as follows: General Shear (the case
upon which the theory is based)
Local Shear Punching Shear Which one takes place
depends upon consistency or density of soil, which decreases from general to local to punching
Generally, with softer soils, settlement tends to govern the design more than bearing capacity
Soil Property Corrections for Local and Punching Shear
● Local and punching shear are accounted for by reducing the cohesion and/or friction angle of the soil
● Read entire section on how and when to use this reduction
● Also analyze for settlement
Bearing Capacity Correction Factors
Shape FactorsShape Factors
Inclined Base Factors
Groundwater Table Correction Factors
Embedment Depth Factors
Load Inclination Factors A convenient way to account for the
effects of an inclined load applied to the footing by the column or wall stem is to consider the effects of the axial and shear components of the inclined load individually. If the vertical component is checked against the available bearing capacity and the shear component is checked against the available sliding resistance, the inclusion of load inclination factors in the bearing capacity equation can generally be omitted. The bearing capacity should, however, be evaluated by using effective footing dimensions, as discussed in Section 8.4.3.1 and in the footnote to Table 8-4, since large moments can frequently be transmitted to bridge foundations by the columns or pier walls. The simultaneous application of shape and load inclination factors can result in an overly conservative design.
Unusual column geometry or loading configurations should be evaluated on a case-by-case basis relative to the foregoing recommendation before the load inclination factors are omitted. An example might be a column that is not aligned normal to the footing bearing surface. In this case, an inclined footing may be considered to offset the effects of the inclined load by providing improved bearing efficiency (see Section 8.4.3.4). Keep in mind that bearing surfaces that are not level may be difficult to construct and inspect. (FHWA NHI-06-089)
Allowable Bearing Capacity
Bearing Capacity Example Given
Square shallow foundation, 5’ x 5’
Foundation depth = 2’ Cohesionless Unit weight 121 pcf Internal friction angle 31
degrees Load on foundation = 76
kips Groundwater table very deep
Find Factor of safety against
bearing capacity failure Solution
Governing equation:
We can neglect factors due to groundwater table (Cw), load inclination (b) and depth (d)
Bearing Capacity Example
Bearing capacity “N” factors for 31 degree friction angle Nc = 32.7 Nq = 20.6 Nγ = 26.0
Bearing Capacity Example● Shape Factors● sc = 1+(5/5)(20.6/32.7)
= 1.63● sγ=1-0.4(5/5) = 0.6● sq = 1+(5/5)(tan(31)) =
1.6
Bearing Capacity Example Other variables
c=0 (problem statement) q=(121)(2) = 242 psf γ= 121 pcf (problem
statement) Substitute and solve
qult = (0) + (242)(20.6)(1.6) + (0.5)(121)(5)(26.0)(0.6) = 0 + 7976.32 + 4719 = 12,695 psf
Compute ultimate load Qult = qult * A (12695)(5)(5) = 317,383
lbs. = 317.3 kips Compute Factor of Safety
FS = 317.3/76 = 4.17 It’s also possible to do this
using the pressures qa = 76,000/(25) = 3040
psf FS = 12,695/3040 = 4.17
Effect of Groundwater Table and Layered Soils on Bearing Capacity
Layered Soils are virtually unavoidable in real geotechnical situations
Softer layers below the surface can and do significantly affect both the bearing capacity and settlement of foundations
Pore water pressure increases; reduces both effective stress and shear strength in the soil (same problem as is experienced with unsupported slopes)
● Three ways to analyze layered soil profiles:
Use the lowest of values of shear strength, friction angle and unit weight below the foundation. Simplest but most conservative. Use groundwater factors in conjunction with this.
Use weighted average of these parameters based on relative thicknesses below the foundation. Best balance of conservatism and computational effort. Use width of foundation B as depth for weighted average
Consider series of trial surfaces beneath the footing and evaluate the stresses on each surface (similar to slope failure analysis.) Most accurate but calculations are tedious; use only when quality of soil data justify the effort
● Groundwater considered using the groundwater correction factor
Groundwater Example● Given
Previous example Groundwater table is 3’ below
base of foundation● Find
Bearing Capacity● Solution
Note that groundwater factor Cwγ is based on Dw, which is distance from surface of soil to groundwater table
Dw = 2’ + 3’ = 5’ Foundation depth plus
distance below the base of the foundation
● Solution Values for interpolation (from Table 8-
5) Dw = Df = 2’ (base of foundation):
Cwq = 1.0 Cwγ = 0.5
Dw = 1.5 * Bf + Df = (1.5)(5’) + 2’ =9.5’ (bottom of influence zone): = 1.0 Cwq = 1.0 Cwγ = 1.0
Interpolating: Cwq = 1.0 Cwγ = 0.7
● Substitute and solve qult = (0) + (242)(20.6)(1.6)(1.0) + (0.5)
(121)(5)(26.0)(0.6)(0.7) = 0 + 7976.32 + 3303.3 = 11,280 psf (11% decrease)
The Other “Workarounds” for Bearing Capacity Theory
● We have already discussed workarounds to the following: Strip footing (shape factors) Level base (base inclination
factors) Vertical load (load
inclination factors) Homogeneous soil
(groundwater factors and related theory)
General Shear (modified c and φ values)
● Other workarounds we will discuss are as follows: Load eccentricity (load is off
centre or there is a moment that accompanies the load Load eccentricity is
unavoidable in some circumstances because of the geometry of the structure and the site
Eccentricity not only impacts bearing capacity but also the basic stability of the foundation base (foundation liftoff)
Footings on top of or on slopes
Eccentric Loading of Foundations● Eccentric loading occurs when
a footing is subjected to eccentric vertical loads, a combination of vertical loads and moments, or moments induced by shear loads transferred to the footing.
Abutments and retaining wall footings are examples of footings subjected to this type of loading condition.
Moments can also be applied to interior column footings due to skewed superstructures, impact loads from vessels or ice, seismic loads, or loading in any sort of continuous frame.
● Eccentricity is the distance from the effective point of loading to the centroid of the foundation.
This distance can be one-way (strip and circular footings) or two-way (square or rectangular footings)
Eccentricity can occur either because the loading is not at the centroid or there is a moment on the foundation.
In the case of a moment, the eccentricity is computed by dividing the moment on the foundation by the applied load, i.e., e = M/P
Ways of Accounting for Eccentricity
Expressing Load Eccentricity and Inclination
Load Divided by Inclination Angle
Total Load P Total Vertical Load Pv Total Horizontal Load Ph Angle of Inclination α =
arctan(Ph/Pv) Load can be concentric or
eccentric Load and Eccentricity
Total Vertical Load P Eccentricity from centroid
of foundation e Horizontal Load (if any)
not included
Moment and Eccentric Load
Total eccentric vertical Load P with eccentricity e
Replace with concentric vertical load P and eccentric moment M=Pe
Continuous Foundations Moments, loads
expressed as per unit length of foundation, thus P/b or M/b
Eccentricity and
Equivalent Footing
Procedure
One Way Loading One-way loading is loading
along one of the centre axes of the foundation
Three cases to consider (see right)
Resultant loads outside the “middle third” result in foundation lift-off and are thus not permitted at all
After this reduced footing size can be computed
Example of One-Way Eccentricity● Given
Continuous Foundation as shown
Groundwater table at great depth
Weight of foundation (concrete) not included in load shown
● Find Whether resultant force
acts in middle third Minimum and maximum
bearing pressures Reduced bearing area
Example of One Way Example of One Way EccentricityEccentricity
Compute Weight of Foundation Wf/b = (5)(1.5)(150) = 1125 lb/ft
Compute eccentricity
Thus, eccentricity is within the “middle third” of the foundation and foundation can be analysed further without enlargement at this point
e=(M /b )Q /b
= 800012000+1125
=0 .61 ft .
B6=5
6=0 .833 ft .>0 .61 ft .
Equations for One-Way Pressures Equations for One-Way Pressures with Eccentric/Moment Loadswith Eccentric/Moment Loads
If q at any point is less than zero, resultant is outside the middle third
Wf is foundation weight
Example of One Way Example of One Way EccentricityEccentricity
Compute minimum and maximum bearing pressures
Example of One-Way Eccentricity● Since the resultant is
within the middle third, we can compute the reduced foundation size B’
● As this is a continuous footing experiencing one-way eccentricity, we do not need to consider an L’
● From previous computations: B = 5’ e = 0.61’
● Reduced Foundation Width B’ = 5 – (2)(0.61) = 3.78’
Two-Way Eccentricity• Eccentricity in both “B” and “L”
directions produces a planar distribution of stress
• Kern of Stability Foundation stable
against overturn only if resultant falls in the kern in the centre of the foundation
Resultant in the kern if6 eBB
+6 eLL
≤1
eB, e
L = eccentricity in B, L directions
Bearing Pressure at CornersBearing Pressure at CornersTwo-Way EccentricityTwo-Way Eccentricity
• Helpful hint to prevent confusion of eccentricity of finite vs. infinite (continuous) foundationso Always use one-way
eccentricity equations for continuous foundations
o Always use two-way eccentricity equations for finite foundations
o Two-way equations will reduce to one-way equations if one of the eccentricities (eB, eL) is zero
L
e
B
e
BL
WQq LBf
6614,3,2,1
Two-Way Eccentricity Example Given
Grain silo design as shown Each silo has an empty
weight of 29 MN; can hold up to 110 MN of grain
Weight of mat = 60 MN Silos can be loaded
independently of each other Find
Whether or not eccentricity will be met with the various loading conditions possible
Eccentricity can be one-way or two-way
Two-Way Eccentricity Two-Way Eccentricity ExampleExample
One-Way Eccentricity Largest Loading: two
adjacent silos full and the rest empty
Q = (4)(29) + 2(110) + 60 = 396 MN
M = (2)(110)(12) = 2640 MN-m
e= MQ
e= 2640396
e=6 .67mB6
=506
=8 .33m>6. 67m
Eccentricity OK for one-way eccentricity
Two-Way Eccentricity Two-Way Eccentricity ExampleExample
Two-Way Eccentricity Largest Loading: one silo full and the rest
empty P = (4)(29) + 110 + 60 = 286 MN MB = ML = (110)(12) = 1320 MN-m
eB=eL=MQ
= 1320286
=4 .62m
6 eBB
+6eLL
=2( ( 6 ) ( 4 .62 )50 )=1 .11>1Not acceptable
Two-Way Eccentricity Two-Way Eccentricity ExampleExample
Two-Way Eccentricity Solution to Eccentricity
Problem: increase the size of the mat
Necessary to also take other considerations into account (bearing failure, settlement, etc.)
6 eBB
+6eLL
=2( ( 6 ) ( 4 .62 )B )=1B=L=55. 4m
Equivalent Footing Using Two-Equivalent Footing Using Two-Way Eccentricity ExampleWay Eccentricity Example
Largest Loading: one silo full and the rest empty
Result of Two-Way Eccentricity Analysis
eB = eL = 4.62 m B = L = 55.4 m (expanded
foundation) Equivalent Footing
Dimensions B’ = B – 2eB = 55.4 – (2)(4.62) B’ = 45.8 m = L’ (as B = L and
eB = eL)
Equivalent Footing Using Two-Equivalent Footing Using Two-Way Eccentricity ExampleWay Eccentricity Example
One-Way Eccentricity Largest Loading: two
adjacent silos full and the rest empty
B = L = 55.4 m (expanded foundation)
eB = 6.67m eL = 0 m B’ = B’-2eB = 55.4 – (2)
(6.67) = 42.1 m L = L’ = 55.4 m B
6=50
6=8 .33m>6. 67m
Other Notes on Bearing Other Notes on Bearing Capacity FactorsCapacity Factors
• Two ways to handle B’ and L’ values when computing shape factors (which are a function of B/L):o AASHTO (2002) guidelines recommend calculating the shape factors, s, by using the effective
footing dimensions, B and L’.′ and L’.o However, the original references (e.g., Vesi , 1975) do not specifically recommend using the ć, 1975) do not specifically recommend using the
effective dimensions to calculate the shape factors. Since the geotechnical engineer typically does not have knowledge of the loads causing eccentricity, use full footing dimensions be used to calculate the shape factors for use in computation of ultimate bearing capacity.
o Either is acceptable for problems in this class. In practice, which one you would use would depend upon a) the project (a highway project would tend to use AASHTO recommendations) and how well the location of the loads was known.
• Bowles (1996) also recommends that the shape and load inclination factors (s and i) should not be combined.
• In certain loading configurations, the designer should be careful in using inclination factors together with shape factors that have been adjusted for eccentricity (Perloff and Baron, 1976). The effect of the inclined loads may already be reflected in the computation of the eccentricity. Thus an overly conservative design may result.
Bearing Capacity for Foundation at Top of a SlopeBearing Capacity for Foundation at Top of a Slope
Example of Footings Example of Footings on Slopeson Slopes
• Solutiono Ncq for D/B = 0 and Slope Angle
of 30 deg. = 4.9
o Ncq for D/B = 1 and Slope Angle of 30 deg. = 6.4
o Linearly interpolating, Ncq =
(6.4+4.9)/2 = 5.7
o Nγq = 1 since the soil is purely cohesive
o B/2 = Do Shape factors are unity because it
is a continuous footing; water table and embedment factors are like wise not considered
– qult = (75)(5.7)+(1.5)(18.5) =
455.25 kPa
• Giveno Strip footing to be constructed on top
of the slopeo Soil properties: c = 75 kPa, γ = 18.5
kN/m3, water table very deepo H = 8 m, B = 3 m, D = 1.5 m, b = 2
m, Slope Angle = 30 deg.
• Findo Ultimate Bearing Capacity of
Footing, using solution method of previous slide
o Solutiono B < H since 3 m < 8 mo Obtain Ncq from Figure 8-18(e) for
Case I with No = 0o D/B = 1.5/3 = 0.5o b/B = 2/3 = 0.667
Required Footing Required Footing SetbacksSetbacks
Practical Aspects of Bearing Capacity Formulations
Failure Zones for Bearing Capacity
Presumptive Bearing CapacityPresumptive Bearing Capacity
Presumptive Bearing Capacity on Rock
Questions?
ENCE 4610 Foundation Analysis and DesignTopics for Shallow Foundations, Part ITypes of Shallow FoundationsFootingsAbutment/Wingwall FootingCombined FootingSlide 7Slide 8Shear Failure vs. Settlement in Allowable Bearing CapacityPlasticity: Lower and Upper Bound SolutionsLower Bound SolutionSlide 12More Realistic Upper Bound CaseDevelopment of Prandtl Bearing Capacity TheorySlide 15Slide 16Basic Equation of Bearing CapacitySlide 18Bearing Capacity ExampleTypes of Bearing CapacitySoil Property Corrections for Local and Punching ShearBearing Capacity Correction FactorsSlide 23Inclined Base FactorsGroundwater Table Correction FactorsEmbedment Depth FactorsLoad Inclination FactorsAllowable Bearing CapacitySlide 29Slide 30Slide 31Slide 32Effect of Groundwater Table and Layered Soils on Bearing CapacityGroundwater ExampleThe Other “Workarounds” for Bearing Capacity TheoryEccentric Loading of FoundationsWays of Accounting for EccentricityExpressing Load Eccentricity and InclinationSlide 39One Way LoadingExample of One-Way EccentricitySlide 42Slide 43Slide 44Slide 45Two-Way EccentricitySlide 47Two-Way Eccentricity ExampleSlide 49Slide 50Slide 51Slide 52Slide 53Slide 54Slide 55Slide 56Slide 57Practical Aspects of Bearing Capacity FormulationsFailure Zones for Bearing CapacitySlide 60Presumptive Bearing Capacity on RockSlide 62